codeforces236B Easy Number Challenge

最新推荐文章于 2021-02-10 00:08:46 发布

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本文介绍了一种计算特定范围内所有整数乘积的约数总和的方法，并提供了两种不同的算法实现，通过模运算确保结果在限定范围内。

Let's denote d(n) as the number of divisors of a positive integer n. You are given three integers a, b and c. Your task is to calculate the following sum:

$\text{[math]}$

Find the sum modulo 1073741824 (2³⁰).

Input

The first line contains three space-separated integers a, b and c (1 ≤ a, b, c ≤ 100).

Output

Print a single integer — the required sum modulo 1073741824 (2³⁰).

Sample test(s)

input
2 2 2

output
20

input
5 6 7

output
1520

Note

For the first example.

d(1·1·1) = d(1) = 1;
d(1·1·2) = d(2) = 2;
d(1·2·1) = d(2) = 2;
d(1·2·2) = d(4) = 3;
d(2·1·1) = d(2) = 2;
d(2·1·2) = d(4) = 3;
d(2·2·1) = d(4) = 3;
d(2·2·2) = d(8) = 4.

So the result is 1 + 2 + 2 + 3 + 2 + 3 + 3 + 4 = 20.

分析：求一个数的约数有几个

#include<cstdio>
#include<cstring>
const int N=1000001;
const int M=1<<30;
int d[N],p[N];
bool f[N];
void get_prime(int n){
	int i,j,cnt=0;f[0]=f[1]=1;
	for(i=2;i<n;i++){
		if(!f[i])p[cnt++]=i;
		for(j=0;j<cnt&&p[j]*i<N;j++){
			f[i*p[j]]=1;
			if(i%p[j]==0)break;
		}
	}
}
void get_prime1(int n){
	int i,j,cnt=0;f[0]=f[1]=1;
	for(i=2;i*i<n;i++){
		if(f[i])continue;
		for(j=i;j*j<n;j++)f[i*j]=1;
	}
	for(i=2;i<N;i++)if(!f[i])p[cnt++]=i;
}
int count(int n){
	if(d[n])return d[n];
	int i=0,a=n,cnt;d[n]=1;
	while(a!=1){
		cnt=0;
		while(a%p[i]==0){
			cnt++;
			a/=p[i];
		}
		if(cnt)d[n]*=(1+cnt);
		i++;
	}
	return d[n];
}
int main(){
	int a,b,c,i,j,k,ans;
	get_prime(N);
	while(~scanf("%d%d%d",&a,&b,&c)){
		ans=0;
		for(i=1;i<=a;i++)for(j=1;j<=b;j++)for(k=1;k<=c;k++){
			ans+=count(i*j*k);if(ans>=M)ans-=M;
		}
		printf("%d\n",ans);
	}
	return 0;
}

方法二：

#include<cstdio>
#include<cstring>
const int N=1000001;
const int M=1<<30;
int d[N];
int main(){
	int a,b,c,i,j,k,ans,J;
	while(~scanf("%d%d%d",&a,&b,&c)){
		ans=0;J=a*b*c;memset(d,0,sizeof(d));
		for(i=1;i<=J;i++)for(j=1;i*j<=J;j++)d[i*j]++;
		for(i=1;i<=a;i++)for(j=1;j<=b;j++)for(k=1;k<=c;k++){
			ans+=d[i*j*k];if(ans>=M)ans-=M;
		}
		printf("%d\n",ans);
	}
	return 0;
}